Homotopy Type Theory polynomial ring > history (Rev #4, changes)

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Definition

Given a commutative ring RR and a type TT, the polynomial ring R[T]R[T] is the free commutative algebra generated by TT, or equivalently, it is a higher inductive type R[T]R[T] inductively generated by

  • a function j:TR[T]j:T \to R[T] representing the function of generators to indeterminants of the polynomial ring
  • a function s:RR[T]s:R \to R[T] representing the function of scalars to constant polynomials of the polynomial ring
  • a term c:isCRing(R[T])c:isCRing(R[T]) representing that the polynomial ring is a commutative ring.
  • a term h:isCRingHom(s)h:isCRingHom(s) representing that s:RR[T]s:R \to R[T] is a commutative ring homomorphism?.

The terma of a polynomial ring are called polynomials.

A term a:R[T]a:R[T] is a constant polynomial if the fiber of ss at aa is inhabited

isConstant(a)fiber(s,a)isConstant(a) \coloneqq \Vert fiber(s, a) \Vert

A term a:R[T]a:R[T] is an indeterminant if the fiber of tt at aa is inhabited

isIndeterminant(a)fiber(t,a)isIndeterminant(a) \coloneqq \Vert fiber(t, a) \Vert

Properties

Partial derivative

Suppose TT has decidable equality? and RR is an integral domain. For every polynomial ring R[T]R[T], one could define a function called a partial derivative

():T(R[T]R[T])\frac{\partial}{\partial(-)}: T \to (R[T] \to R[T])

such that there are dependent terms

t:Tλ(t): a:R[T] b:R[T]isConstant(a)×isConstant(b)× p:R[T] q:R[T]t(ap+bq)=at(p)+bt(q)t:T \vdash \lambda(t):\prod_{a:R[T]} \prod_{b:R[T]} isConstant(a) \times isConstant(b) \times \prod_{p:R[T]} \prod_{q:R[T]} \frac{\partial}{\partial t}(a \cdot p + b \cdot q) = a \cdot \frac{\partial}{\partial t}(p) + b \cdot \frac{\partial}{\partial t}(q)
t:Tπ(t): p:R[T] q:R[T]t(pq)=pt(q)+t(p)qt:T \vdash \pi(t): \prod_{p:R[T]} \prod_{q:R[T]} \frac{\partial}{\partial t}(p \cdot q) = p \cdot \frac{\partial}{\partial t}(q) + \frac{\partial}{\partial t}(p) \cdot q
t:Tι(t): u:R[T]isIndeterminant(u)×(u=t)×(t(u)=1)t:T \vdash \iota(t): \prod_{u:R[T]} \isIndeterminant(u) \times (u = t) \times \left(\frac{\partial}{\partial t}(u) = 1\right)
t:Tc(t): u:R[T]isIndeterminant(u)×(ut)×(t(u)=0)t:T \vdash c(t): \prod_{u:R[T]} \isIndeterminant(u) \times (u \neq t) \times \left(\frac{\partial}{\partial t}(u) = 0\right)

Thus the polynomial ring R[T]R[T] is a differential algebra?.

See also

Revision on June 17, 2022 at 14:17:37 by Anonymous?. See the history of this page for a list of all contributions to it.